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$f:X\mapsto Y$ is uniformly continuous. $X \subset T$ (T is a complete metric space e.g.$T=\mathbb{R}$). Define $X' = \{x:x \text{ is a limit point of }X\}$. For any $p \in X'\backslash X$, does $\lim\limits_{x \to p} f(x) $ exist?

Intuitively I think the limit exists but I cannot prove it.

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  • $\begingroup$ As $X'\subseteq X$, $X'\setminus X$ is empty. Is there a typo? $\endgroup$ – James May 7 '16 at 21:04
  • $\begingroup$ @James $X'$ is not a subset of $X$, for example, $X=(0,1), X'=[0,1], X'\backslash X = \{0,1\}$ $\endgroup$ – Huayu Zhang May 7 '16 at 21:06
  • $\begingroup$ Oh, then you need some ambient space $T$ in which $X$ is living. $\endgroup$ – James May 7 '16 at 21:07
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No wonder you cannot, since it is not true in general. Example: $$ X=(0,1)\subset\mathbb R,\quad Y=\mathbb R,\quad f(x)=\sin\dfrac1x. $$

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  • $\begingroup$ What if the function is uniformly continuous? $\endgroup$ – Huayu Zhang May 7 '16 at 21:12
  • $\begingroup$ Then it is true if $X$ is a subset of a complete metric space. You prove it showing that if $x_n\to p$, then $f(x_n)$ is a Cauchy sequence. $\endgroup$ – Julián Aguirre May 7 '16 at 21:16
  • $\begingroup$ I need to show for any sequence $\{x_n\}$ if $x_n \to p$, then $f(x_n)$ converges to a fixed point $y$. I can show all $f(x_n)$ are Cauchy sequences, but this does not guarantee they will converge to the same $y$. $\endgroup$ – Huayu Zhang May 7 '16 at 21:28
  • $\begingroup$ You can prove that the limit is unique. If $x_n$ and $x'_n$ converge to $p$, then $f(x_n)-f(x'_n)$ is small for $n$ large enough. $\endgroup$ – Julián Aguirre May 7 '16 at 21:36

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